$f(x) = \begin{cases} 7 & \text{if } x = 0 \\ 2x^{2}+1 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Explanation: First consider the behavior for $x \ne 0$ Consider the range of $2x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $2$ doesn't change the range. To get $2x^{2}+1$ , we add $1$ So the range becomes: $\{\, y \mid y ≥ 1 \,\}$ If $x = 0$ , then $f(x) = 7$ , which eliminates $f(x) = 1$ from the range. The new range is $\{\, y \mid y > 1 \,\}$.